Multiplication Theorem and Independent Events

Multiplication Theorem of Probability

The multiplication theorem follows directly from the definition of conditional probability. It helps us find the probability of simultaneous occurrence of two or more events.

Theorem for Two Events

If $A$ and $B$ are any two events and $P(A) \neq 0$, then $$P(A \cap B) = P(A) \cdot P(B|A)$$ Similarly, if $P(B) \neq 0$, then $$P(A \cap B) = P(B) \cdot P(A|B)$$

Theorem for Three Events

If $A$, $B$, and $C$ are three events with the relevant conditional probabilities defined, then $$P(A \cap B \cap C) = P(A) \cdot P(B|A) \cdot P(C|A \cap B)$$ This is an important special case of the chain rule of probability.

Independent Events

Two events are said to be independent if the occurrence or non-occurrence of one does not affect the probability of occurrence of the other.

Equivalent Conditions

If $P(A) \neq 0$ and $P(B) \neq 0$, then the following are equivalent: $$P(A|B) = P(A)$$ $$P(B|A) = P(B)$$ $$P(A \cap B) = P(A)\,P(B)$$

Multiplication Rule for Independent Events

If $A$ and $B$ are independent, then $$P(A \cap B) = P(A)\,P(B)$$

Extension to More Than Two Events

Events $A_1, A_2, \dots, A_n$ are said to be mutually independent if the probability of the intersection of every collection of these events equals the product of their respective probabilities. In particular, $$P(A_1 \cap A_2 \cap \dots \cap A_n) = P(A_1)P(A_2)\cdots P(A_n)$$

Mutually Exclusive vs. Independent Events

This is a very common source of confusion.

• Mutually Exclusive Events: They cannot occur together. Thus, $$A \cap B = \varnothing \implies P(A \cap B)=0$$ If $P(A)>0$ and $P(B)>0$, then such events cannot be independent.

• Independent Events: They may occur together, but the occurrence of one does not affect the probability of the other.

Key Fact: If $A$ and $B$ are independent, then the pairs $(A, B')$, $(A', B)$, and $(A', B')$ are also independent.

Example 1: Drawing Cards Without Replacement

Two cards are drawn from a well-shuffled deck of 52 cards, one after another without replacement. Find the probability that both cards are red.

Solution: Let $$A = \text{event that the first card is red}, \qquad B = \text{event that the second card is red}$$

For the first draw, $$P(A) = \frac{26}{52} = \frac{1}{2}$$

Given that the first card is red, there remain $25$ red cards out of $51$ cards. Hence, $$P(B|A) = \frac{25}{51}$$

By the multiplication theorem, $$P(A \cap B) = P(A) \cdot P(B|A)$$ $$P(A \cap B) = \frac{1}{2} \cdot \frac{25}{51} = \frac{25}{102}$$

Answer: $$\boxed{\frac{25}{102}}$$

Example 2: Checking for Independence

A die is thrown. If $E$ is the event 'the number appearing is a multiple of 3' and $F$ is the event 'the number appearing is even', find whether $E$ and $F$ are independent.

Solution: The sample space is $$S = \{1,2,3,4,5,6\}$$

Now, $$E = \{3,6\}, \qquad F = \{2,4,6\}$$

Therefore, $$P(E) = \frac{2}{6} = \frac{1}{3}$$ $$P(F) = \frac{3}{6} = \frac{1}{2}$$

Also, $$E \cap F = \{6\}$$ so $$P(E \cap F) = \frac{1}{6}$$

Now check: $$P(E)P(F) = \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6}$$

Since $$P(E \cap F) = P(E)P(F)$$ we conclude that $E$ and $F$ are independent.

Answer: $$\boxed{\text{The events } E \text{ and } F \text{ are independent.}}$$

Example 3: Solving a Problem Independently

Probability of solving a specific problem independently by $A$ and $B$ are $\frac{1}{2}$ and $\frac{1}{3}$ respectively. If both try to solve the problem independently, find the probability that the problem is solved.

Solution: Let $$A = \text{event that person } A \text{ solves the problem}$$ $$B = \text{event that person } B \text{ solves the problem}$$

Given, $$P(A)=\frac{1}{2}, \qquad P(B)=\frac{1}{3}$$

The problem is solved if at least one of them solves it, i.e. $$P(A \cup B)$$

Since they work independently, $$P(A \cap B)=P(A)P(B)=\frac{1}{2}\cdot\frac{1}{3}=\frac{1}{6}$$

Now, $$P(A \cup B)=P(A)+P(B)-P(A \cap B)$$ $$P(A \cup B)=\frac{1}{2}+\frac{1}{3}-\frac{1}{6}$$ $$P(A \cup B)=\frac{3+2-1}{6}=\frac{4}{6}=\frac{2}{3}$$

Answer: $$\boxed{\frac{2}{3}}$$

Example 4: Drawing Marbles from an Urn

An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other without replacement. What is the probability that both drawn balls are black?

Solution: Let $$B_1 = \text{event that the first ball is black}, \qquad B_2 = \text{event that the second ball is black}$$

Then, $$P(B_1)=\frac{10}{15}=\frac{2}{3}$$

After one black ball is drawn, there remain $9$ black balls out of $14$ total balls. Hence, $$P(B_2|B_1)=\frac{9}{14}$$

By multiplication theorem, $$P(B_1 \cap B_2)=P(B_1)\cdot P(B_2|B_1)$$ $$P(B_1 \cap B_2)=\frac{2}{3}\cdot\frac{9}{14}=\frac{18}{42}=\frac{3}{7}$$

Answer: $$\boxed{\frac{3}{7}}$$

Example 5: Three Independent Events

A, B, and C shoot at a target. Their probabilities of hitting the target are $0.4, 0.5$, and $0.7$ respectively. Find the probability that at least one of them hits the target.

Solution: Let the events $A$, $B$, and $C$ denote hitting the target. Then the probabilities of missing are: $$P(A')=1-0.4=0.6$$ $$P(B')=1-0.5=0.5$$ $$P(C')=1-0.7=0.3$$

Since the trials are independent, $$P(\text{none hits})=P(A' \cap B' \cap C')=P(A')P(B')P(C')$$ $$P(\text{none hits})=0.6\times0.5\times0.3=0.09$$

Therefore, $$P(\text{at least one hits})=1-P(\text{none hits})$$ $$P(\text{at least one hits})=1-0.09=0.91$$

Answer: $$\boxed{0.91}$$

Example 6: Cards and Independence

In a deck of 52 cards, let $E$ be the event 'the card is a Spade' and $F$ be the event 'the card is an Ace'. Are $E$ and $F$ independent?

Solution: There are $13$ spades in the deck, so $$P(E)=\frac{13}{52}=\frac{1}{4}$$

There are $4$ aces in the deck, so $$P(F)=\frac{4}{52}=\frac{1}{13}$$

The event $E \cap F$ means the card is the Ace of Spades. There is exactly one such card. Hence, $$P(E \cap F)=\frac{1}{52}$$

Now, $$P(E)P(F)=\frac{1}{4}\cdot\frac{1}{13}=\frac{1}{52}$$

Since $$P(E \cap F)=P(E)P(F)$$ we conclude that $E$ and $F$ are independent.

Answer: $$\boxed{\text{Yes, } E \text{ and } F \text{ are independent.}}$$

Example 7: Exactly One of Two Events Occurs

If $A$ and $B$ are independent events with $P(A) = 0.3$ and $P(B) = 0.4$, find the probability that exactly one of them occurs.

Solution: Exactly one of $A$ and $B$ occurs means either $$A \cap B'$$ or $$A' \cap B$$ These two events are mutually exclusive. Therefore, $$P(\text{exactly one})=P(A \cap B')+P(A' \cap B)$$

Since $A$ and $B$ are independent, we also have independence of complements. Thus, $$P(A \cap B')=P(A)P(B')=0.3\times(1-0.4)=0.3\times0.6=0.18$$ $$P(A' \cap B)=P(A')P(B)=(1-0.3)\times0.4=0.7\times0.4=0.28$$

So, $$P(\text{exactly one})=0.18+0.28=0.46$$

Answer: $$\boxed{0.46}$$

Example 8: Probability of Intersection Using Independence

If $A$ and $B$ are two independent events such that $P(A \cup B) = 0.6$ and $P(A) = 0.2$, find $P(B)$.

Solution: Let $$P(B)=p$$ Since $A$ and $B$ are independent, $$P(A \cap B)=P(A)P(B)=0.2p$$

Using the addition theorem, $$P(A \cup B)=P(A)+P(B)-P(A \cap B)$$ Substitute the given values: $$0.6=0.2+p-0.2p$$ $$0.4=0.8p$$ $$p=\frac{0.4}{0.8}=0.5$$

Therefore, $$P(B)=0.5$$

Answer: $$\boxed{0.5}$$

Example 9: Drawing with Replacement

A box contains 10 red and 8 blue marbles. Two marbles are drawn one by one with replacement. Find the probability that the first is red and the second is blue.

Solution: Since the draws are made with replacement, the two events are independent.

Probability that the first marble is red: $$P(R_1)=\frac{10}{18}=\frac{5}{9}$$

Probability that the second marble is blue: $$P(B_2)=\frac{8}{18}=\frac{4}{9}$$

Therefore, $$P(R_1 \cap B_2)=P(R_1)P(B_2)$$ $$P(R_1 \cap B_2)=\frac{5}{9}\cdot\frac{4}{9}=\frac{20}{81}$$

Answer: $$\boxed{\frac{20}{81}}$$

Example 10: Logic of Complements

Prove that if $A$ and $B$ are independent, then $A$ and $B'$ are also independent.

Solution: Since $$A = (A \cap B) \cup (A \cap B')$$ and the two events on the right-hand side are mutually exclusive, we have $$P(A)=P(A \cap B)+P(A \cap B')$$ Thus, $$P(A \cap B')=P(A)-P(A \cap B)$$

Because $A$ and $B$ are independent, $$P(A \cap B)=P(A)P(B)$$ So, $$P(A \cap B')=P(A)-P(A)P(B)$$ $$P(A \cap B')=P(A)\bigl(1-P(B)\bigr)$$ But $$1-P(B)=P(B')$$ Hence, $$P(A \cap B')=P(A)P(B')$$ Therefore, $A$ and $B'$ are independent.

Answer: $$\boxed{\text{Hence proved that } A \text{ and } B' \text{ are independent.}}$$